I'm trying to set a path to an icon in my project.
"/icons/currency/USD.png"
But I'm getting an error:
FileNotFoundError: [Errno 2] No such file or directory: 'icons/currency/USD.png'
If I set the path to "/icons/USD.png"
everything is work.
but I don't want all my icons will be in one folder so I open currency folder
to separate it from other icons.
my project folders structure is:
project folder -> name_of_project.py , icons folder
icons folder -> currency folder -> USD.png
What is the correct way to configure this relative path ?
I'd recommend using os.path.join() to get full paths without bothering with your OS's specifics.
Currently you are setting your path to start from the root of your OS (I'm assuming Linux because of the path you have given as example). What you need is to get the path (absolute) of the file you are running and navigating from there.
You can do it as follows:
abs_path_of_executable_file = os.path.split(os.path.abspath(__file__))[0]
path_to_png = os.path.join(abs_path_of_executable_file, 'icons', 'currency', 'USD.png')
# do stuff with path_to_png
Where the first line gets the absolute path of the file you are executing (os.path.abspath(__file__)) and then you are taking the full path of the directory containing that file (os.path.split('some/path/to/file.py')[0]).
os.path.split('some/path/to/file.py')[1] gives you only the filename, without any paths.
Related
I have saved a file. I have not given a path to this. In the project folder, nowhere can I find the file.
Is there any method that will tell me where the default path is where the file has gone?
model_file = "test.h5"
model.save(model_file, overwrite=True)
print(os.getcwd()) will print the working directory, where relative paths start from (and where a relative filename such as foo.txt will end up in).
I am getting a FileNotFound error when iterating through a list of files in Python on Windows.
The specific error I get looks like:
FileNotFoundError: File b'fileName.csv' does not exist
In my code, I first ask for input on where the file is located and generate a list using os (though I also tried glob):
directory = input('In what directory are your files located?')
fileList = [s for s in os.listdir(directory) if s.endswith('.csv')]
When I print the list, it does not contain byte b before any strings, as expected (but I still checked). My code seems to break at this step, which generates the error:
for file in fileList:
pd.read_csv(file) # breaks at this line
I have tried everything I could find on Stack Overflow to solve this problem. That includes:
Putting an r or b or rb right before the path string
Using both the relative and absolute file paths
Trying different variations of path separators (/, \, \\, etc.)
I've been dealing with Windows-related issues lately (since I normally work in Mac or Linux) so that was my first suspicion. I'd love another set of eyes to help me figure out where the snag is.
A2A.
Although the list was generated correctly because the full directory path was used, the working directory when the file was being run was .. You can verify this by running os.getcwd(). When later iterating through the list of file names, the program could not find those file names in the . directory, as it shouldn't. That list is just a list of file names; there was no directory tied to it so it used the current directory.
The easiest fix here is to change the directory the file is running through after the input. So,
directory = input('In what directory are your files located?')
os.chdir(directory) # Points os to given directory to avoid byte issue
If you need to access multiple directories, you could either do this right before you switch to each directory or save the full file path into your list instead.
I am so new with python and pycharm and i got confuse!!
When I run my project in pycharm it gives me an error about not finding the path of my file. The physical file path is:
'../Project/BC/RequiredFiles/resources/a_reqs.csv'
My project working directory is "Project/BC" and the project running file (startApp.sh) is there too. but the .py file that wants to work with a_req.csv is inside the "RequiredFiles" folder. There is the following code in the .py file:
reqsfile = os.getcwd() + "/resources/a_reqs.csv"
it returns: '../Project/BC/resources/a_reqs.csv'
instead of: '../Project/BC/resources/RequiredFiles/a_reqs.csv'
while the .py file is in "RequiredFiles" the os.getcwd() must include it too. but it does not.
The problem is that i can not change the addressing code. because this code works in another IDE and other people who work with the code in other platform or OS do not have any problem. I am working in mac OS and if i am not mistaken the code works with windows!!
So, how can i tell Pycharm (in mac) to see and load "RequiredFiles" folder as the subfolder of my working directory!!!
os.getcwd returns the current working directory of the process (which may be the directory where startApp.sh is located or another one, depending on the PyCharm's run configuration setting, or, if you start the program from the command line, the directory in which you execute the command).
To make a path independent on the current working directory, you can take the directory where your Python file is located and build the path from it:
os.path.dirname(__file__) + "/resources/a_reqs.csv"
From your question what I see is:
reqsfile = os.getcwd() + "/resources/a_reqs.csv"
Which produces: "../Project/BC/resources/a_reqs.csv", whereas your desired output is
"../Project/BC/resources/RequiredFiles/a_reqs.csv". Since we know os.getcwd is returning "/Project/BC/", then to get your desired result you should be doing:
reqsfile = os.getcwd() + "/resources/RequiredFiles/a_reqs.csv"
But since you want the solution to work with or without the RequiredFiles subdirectory you could apply a conditional solution, ie something like:
import os.path
if os.path.exists(os.getcwd() + "/resources/RequiredFiles/a_reqs.csv"):
reqsfile = os.getcwd() + "/resources/RequiredFiles/a_reqs.csv"
else:
reqsfile = os.getcwd() + "/resources/a_reqs.csv"
This solution will set the reqsfile to the csv in the RequiredFiles directory if the directory exists, and thus will work for you. On the other-hand, if the RequiredFiles directory doesn't exist, it will default to the csv in /resources/.
Typically when groups collaborate on projects, the maintain the same file hierarchy so that these types of issues are avoided, so you might want to consider moving the csv from /RequiredFiles/ to /resources/.
I am using python on mac and would like to open a pdf file that is present in different directory than the directory my main python code is running. I tried different options but there is always an error saying file doesn't exist even when the file is present or [Error no. 2] file cannot be opened. Here is the code that I use:
helpFile = os.path.abspath('~/help/help.pdf')
self.help_btn = tk.Button(self.help_frm, text="Help!", width=8, command = lambda: os.system("open "+helpFile))
could some one help please.
abspath does not expand ~ into the user's home directory, it just calculates the absolute path of a file based on its path relative to the current working directory.
From the docs, it is equivalent to:
normpath(join(os.getcwd(), path))
So in your code, helpFile is being set to "/path/to/cwd/~/help/help.pdf"
To expand ~, use os.path.expanduser.
This question already has answers here:
How do I get the parent directory in Python?
(21 answers)
Closed 9 years ago.
So I've just coded this class for a title screen and it works well. However, one of the people I'm working with on the project mentioned that I shouldn't use:
os.chdir(os.getcwd() + "/..")
resource = (os.getcwd() + "/media/file name")
to get to the super directory. He did mention something about the pythonpath though. We're using Eclipse if this is of some help.
For more context we're making a multi-platform game so we can't just synchronize our directories and hard-code it (although we are using git so the working directory is synchronized). Basically, I need some way to get from a script file in a "src' folder to a "media" folder that's next to it (AKA There's a super (project) folder with both "src" and "media" folders in it).
Any help would be greatly appreciated, but please don't say "google it" because I tried that before coming here (I don't know if that's a frequent thing here, but I've seen it too many times elsewhere...when I've googled for answers, sorry if I sound jerkish for saying that)
Python programs do have the concept of a current working directory, which is generally the directory from which the script was run. This is "where they look for files" with a relative path.
However, since your program can be run from a different folder than the one it is in, your directory of reference needs to instead refer to the directory your script is in (the current directory is not related to the location of your script, in general). The directory where your script is found is obtained with
script_dir = os.path.dirname(__file__)
Note that this path can be relative (possibly empty), so it is still important that the current working directory of your script be the same as the directory when your script was read by the python interpreter (which is when __file__ is set). It is important to convert the possibly relative script_dir into an absolute path if the current working directory is changed later in the code:
# If script_dir is relative, the current working directory is used, here. This is correct if the current
# working directory is the same as when the script was read by the Python interpreter (which is
# when __file__ was set):
script_dir = os.path.abspath(script_dir)
You can then get to the directory media in the parent directory with the platform-independent
os.path.join(script_dir, os.path.pardir, 'media')
In fact os.path.pardir (or equivalently os.pardir) is the platform-independent parent directory convention, and os.path.join() simply joins paths in a platform independent way.
I'd recommend something like:
import os.path
base_folder = os.path.dirname(os.path.dirname(os.path.abspath(__file__)))
media_folder = os.path.join(base_folder, "media")
src_folder = os.path.join(base_folder, "src")
resource = os.path.join(media_folder, "filename")
for path in [base_folder, media_folder, src_folder, resource]:
print path
The main ingredients are:
__file__: gets the path to the current source file (unlike sys.argv[0], which gives the path the script that was called)
os.path.split(): splits a path into the relative file/folder name and the base folder containing it. Using it twice as in base_folder = ... will give the parent directory.
os.path.join: OS-independent and correct combination of path names. Is aware of missing or multiple /s or \s
Consider using os.path.dirname() and os.path.join(). These should work in a platform independent way.